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Calculate pH at which Mg(OH)(2) begins t...

Calculate pH at which `Mg(OH)_(2)` begins to precipitate from a solution containing `0.10M Mg^(2+)` ions. `(K_(SP)of Mg(OH)_(2)=1xx10^(-11))`

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`[Mg^(2+)] =0.1 M " and " K_(sp) " for " Mg(OH)_(2) =1.0 xx 10^(-11)`
the equilibrium in the saturated solution
`Mg(OH)_(2) hArr Mg^(2+) + 2OH^(-)`
`[OH^(-)]^(2) =(K_(sp))/([Mg^(2+)]] =(1.0 xx 10^(-11))/(0.1) =1.0 xx 10^(-10)`
`:. [OH]^(-) =(1.0 xx 10^(-10))^(1//2) =10^(-5) M`
We know that `[H_(3)O^(+)][OH^(-)] =K_(w) = 10^(-14) :. [H_(3)O^(+)] = (10^(-14))/([OH^(-)]] =(10^(-14))/(10^(-5)) =10^(-9) M`
`pH =- log [H_(3)O^(+)] =- log [10^(-9)] =(-) (-9) log 10 =9`
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